20Number 20ExplanationIf α\alphaα and β\betaβ are roots of the equation log3x−logx(2x−4+4x)=1\log_3 x - \log_x\left(2x - 4 + \frac{4}{x}\right) = 1log3x−logx(2x−4+x4)=1, then α+β=....\alpha + \beta = ....α+β=....−2-2−2−1-1−1000222444ExplanationTransform the equation to exponential form logx3−logx(2x−4+4x)=1\log_x 3 - \log_x\left(2x - 4 + \frac{4}{x}\right) = 1logx3−logx(2x−4+x4)=1logx32x−4+4x=logxx\log_x \frac{3}{2x - 4 + \frac{4}{x}} = \log_x xlogx2x−4+x43=logxx32x−4+4x=x\frac{3}{2x - 4 + \frac{4}{x}} = x2x−4+x43=x32x−4+4x=x1\frac{3}{2x - 4 + \frac{4}{x}} = \frac{x}{1}2x−4+x43=1x(2x−4+4x)x=3\left(2x - 4 + \frac{4}{x}\right)x = 3(2x−4+x4)x=32x2−4x+4=32x^2 - 4x + 4 = 32x2−4x+4=32x2−4x+1=02x^2 - 4x + 1 = 02x2−4x+1=0 Therefore the sum of the roots is α+β=−ba=−(−4)2=2\alpha + \beta = \frac{-b}{a} = \frac{-(-4)}{2} = 2α+β=a−b=2−(−4)=2
